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how to differentiate f(x) = ln ( 1 + √ x / 1 - √ x )

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In general, \[ \frac{d}{dx} \ln( f(x) ) = \frac{1}{f(x)}\cdot f'(x)\]
how do i solve it? i cant seem to get the correct answer.
Hint: You will have to use the chain rule by finding the derivative of the argument and multiplying it by the derivative that results without the chain rule. Are you aware of how to use the Chain Rule? If not, then I'll provide a solution.

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im aware of it. but how do i calculate it?
The derivative of the argument, you mean?
how do i apply chain rule in this equation
Rewrite f(x): \(\Large f(x)=ln\frac{1+\sqrt x}{1-\sqrt x}=ln(1+\sqrt x)-ln(1-\sqrt x) \) now just take the derivatives of each ln separately....
Would you mind showing us your work? What have you found for the derivative of the argument of the natural log?
dy/dx = [ (1/1+√ x) (+ 1/2√x) ] - [ (1/1-√x) - ( 1/2√ x) ]

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